hiper-nitrat de ferro 3

hiper-nitrat de Ferro 3:


2·N(OH)_{3} + Fe_{2} + e^{-} ==> 2·HNFe + 3·H_{2}O_{2}
2·HNFe + 3·H_{2}O_{2} + e^{+} ==> 2·N(OH)_{3} + Fe_{2}


Entalpia de reacción:
[ 2·HN(OH)_{3} ] = 18
[ Fe_{2} ] = 6
[ 2·NFe ] = 12
[ 3·H_{2}O_{2} ] = 12


18+6 = 12+12 = 24